Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the surface area of the hyperbolic paraboloid $z = 13 + x^2 - y^2$ that lies above the closed disk $x^2 + y^2 \le 4$.

I know that the surface area is equal to: $\iint$$\sqrt{1+[f_x(x,y)]^2 + [f_y(x,y)]^2}$ $dA$ over some region $R$.

$f_x(x,y)$ = $2x$
$f_y(x,y)$ = $-2y$

It's now equal to:

$\iint$$\sqrt{1+4x^2 + 4y^2}$ $dA$ for some region $R$

I'm stuck at this point. Can someone help me finish this? Thank you!

share|improve this question
    
Put $r=x^2+y^2$ and replace $dA$ by $r dr d\theta$. –  coffeemath Dec 11 '12 at 11:55
    
@coffeemath At what certain values should I evaluate the double integral? –  Marcus Dec 11 '12 at 12:12

1 Answer 1

up vote 1 down vote accepted

Square roots in integrals generally make problems difficult, but here it's a clue that the problem can be simplified by making a change of variable to polar coordinates. Let $x = r \cos \theta$, $y=r\sin\theta$.

To get you started, we write the region of integration in polar coordinates. The closed disk $x^2+y^2\leq 4$ is given in polar coordinates by $\{(r,\theta)|0\leq r\leq 2, 0 \leq \theta < 2\pi\}$. This gives you very easy limits of integration.

You've already found the integrand in terms of $x$ and $y$. Use the polar coordinate transformation to express it in terms of $r$ and $\theta$, and use trigonometric identities to make the integrand very simple.

Finally, remember that whenever you make a change in variable you must multiply by the appropriate Jacobian determinant. (It's easy to forget this step. Remember that $dA=dxdy$ in our original coordinates, but $dA\neq drd\theta$ in our new coordinates - something's missing.)

share|improve this answer
    
Thank you very much! :D –  Marcus Dec 11 '12 at 15:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.